These notes are intended as a guide for the computer program "Stress, Mohr Circle" which illustrates the use of stress calculations for two-dimensional systems. The program consists of five displays. In each display you can vary the physical situation and immediately see the effects in the graphical displays. The first display illustrates the relation between forces and stresses acting on a rectangular solid. In the second display, you explore the stress on a plane as you vary the orientation of the plane and the state of stress in the system. We illustrate the calculation of the stress on the plane in terms of a simple matrix multiplication in the third display. You can also observe the simple geometry of the envelope of the stresses that are produced as the angle of the plane is varied. The fourth display shows the behavior of the normal and shear components of the stress on the plane. When the normal component is plotted along x and the shear component along y, the envelope will be a circle: called a Mohr circle. This beautiful result lets you visualize the range of possible stresses on the plane. Finally in the last display, you can explore the conditions which lead to the fracture of a sample: you can vary the stresses until the stresses are just strong enough to cause fracture by watching the changes in size and location of the Mohr circle.
Display 1: "Equilibrium"
Stress and force vectors acting on a rectangular solid
A mechanical system deforms or fractures when forces are applied and it is the force per unit area that is the quantity which determines the deformation or failure. There is unfortunately no single name for this concept with the words stress, stress vector, or traction used by various authors all to name the force per unit area acting on a plane. We will use the phrase stress vector. It turns out that the stress vector depends in direction and magnitude on the orientation of the plane on which the force acts. A small array of numbers called a stress tensor allows us to calculate the stress vector for any orientation of the plane. In two dimensions, the stress tensor is a two by two array and three numbers define the stress tensor and allow us to calculate the stress vector for any plane. In three dimensions we must use a three by three array: and six numbers then are sufficient to specify the stress tensor. We will restrict ourselves to two dimensions in these notes. This keeps the calculations and illustrations simple.
We can illustrate the constant force approximation by considering a small volume and dividing the bottom into three equal parts. Then the three forces Ba, Bb, Bc acting on the parts of the bottom are all equal.
The stress vector acting on the bottom is then just the total force B divided by the total area of the bottom and we denote it by s B. With B=Ba+Bb+Bc the stress vector on the bottom is s B =B/(w*d) and with our choice of d=1 this is sB=B/w. In our program, we specify the plane on which the stress vector acts by the normal to that plane. We choose the normal to point into the body and the stress vector is the force per unit area acting onto the plane from the outside of the body. Thus for the force B acting on the bottom surface from below we choose the normal pointing up into our volume, the y axis in this case, and we call the bottom the y plane.
In general the stress vector will not be perpendicular to the plane on which it acts. For the (bottom) y plane shown, the total stress vector consists of two parts: s yx parallel to the x direction and s yy parallel to the y direction. The first subscript (y) gives us the direction of the normal and the second subscript gives the direction for the part of the stress vector. The total stress vector sB on the bottom is just the vector sum of these two parts.
s B = B /w*d=s yx +s yy
We will often want to specify length and direction of the x and y parts by giving just two numbers or components and we will use s yx and s yy for the x and y components of the stress vector acting on the y plane. In the above illustration both components s yx and s yy are positive numbers .
Similarly for the left hand side of area h*d=h*1=h we use L to for the total force and the normal to the left side is just the x axis. So the parts of the stress vector sL acting on the left side are just s xx and s xy. The first subscript x again gives the direction of the normal to the left side and the second subscript gives the direction for the part of the stress vector. In the illustration below and in the first computer display, we show the body with the force vectors on the left of the screen and the body with the stress vectors on the right of the screen. The force vectors on the four sides of the rectangle are L, B, R, and T (left, bottom, right, and top).
The vanishing of the total force means that the force B on the bottom is balanced by T the force on the top and the force L on the left side is balanced by R the force on the right with T=-B and R=-L.
Since the rectangle must not undergo any rotational acceleration around the center, we must have no torque around the center. The force Bx then has a torque Bx*h/2 about the center (just the magnitude of the force multiplied by its perpendicular lever arm) and this torque is counterclockwise for positive Bx. The By part of the force on the bottom exerts no torque around the center since it points at the center and hence has zero perpendicular lever arm. There is an equal torque and in the same sense from T the force on the top since by force balance Tx=-Bx. The torque from the bottom and the top is then 2*Bx*h/2=Bx*h. In the same way the torque from the left and right side is 2*Ly*w/2 =Ly*w but the twist is clockwise for Ly positive. The total counter- clockwise torque is then Bx*h-Ly*w and this must be zero for equilibrium.
Hence Bx*h=Ly*w or dividing by h*w Bx/w=Ly/h. We recognize that Bx/w is just s yx the x component of the stress vector on the bottom and Ly/h is just s xy the y component of the stress vector on the left side. We are therefore left with the important result that s yx=s xy. (The same argument can be used in three dimensions to show that not only s xy=s yx but also s xz=s zx and s yz=s zy). We can summarize our results for the stress vectors acting on a rectangular system:
You should explore the relationship between the stress vectors and the force vector using the first or "equilibrium " display in the program. Appendix 1.1 tells you how to navigate in the program. Use display1 to work exercises 1.1, 1.2, 1.3, and 1.4.
Display 2: "Stress Vector Part 1"
Stress vector on a plane whose normal is at angle a with the x axis
In many real situations it is important to be able to calculate the stress vector acting on a plane which may have an arbitrary orientation with respect to our coordinate axes. Such calculations are crucial in predicting the plane along which an object may fracture under stress. If we can calculate the stress vector for an arbitrarily chosen plane we can say that we know the state of stress of the system. In this section and in the accompanying display 2, you can see how to obtain the stress vector on a plane whose normal makes an arbitrary angle a with respect to the x axis from the three components s xx, s xy, and s yy. These three numbers therefore define the state of stress in two dimensions.
2.1 Equilibrium for a triangular solid
We will consider a small triangular solid with one side parallel to x, the second parallel to y, and the third side perpendicular to a normal making an angle a with respect to the x axis as shown. To simplify our calculations we choose the length l for the third side equal to one. Then the width w of the top of the triangle is w=l*sin(a )=sin(a ) and the height h of the right side is h=1*cos(a )=cos(a ). We’ll call the stress vector acting on the top s T, and the stress vector acting on the right sR. The stress vector sP which acts from the outside onto the third side will be found by finding the force FP on the third side of the triangle. The vector sum of all the forces must vanish for equilibrium: we must have
R+T+FP=0 or FP=-R-T. Because we chose l=1, the area of the third side is l*d=1*1=1 and sP=FP=-R-T.
We want to write the x and y components of sP in terms of the stress components s xx, s yy, ands xy so we will need the x and y components of R and T in terms of the stress components.
Rx=-s xx*h=-s xx*cos(a ), Ry=-s xy*h=-s xy *cos(a ), Tx=-s yx*w=-s yx*sin(a ), and Ty=-s yy*w=-s yy*sin(a ). Using these components in sP=-R-T.
s Px=-Rx-Tx==s xx*cos(a )+s yx*sin(a ),
s Py=-Ry-Ty==s xy *cos(a )+s yy*sin(a ),
These two equations therefore let you calculate the x and y components of the stress vector sP for a plane whose normal makes an arbitrary angle a with respect to the x axis. (The three dimensional versions are often called the Cauchy equations.) In the two dimensional case, the two equations let you calculate the stress vectors for any arbitrary plane This shows that the components s xx, s yy, and s xy completely specify the stress for a two dimensional system. The vectors PN, sP, sPx and sPy are shown in the next figure.
You should now work exercises 2.1,2.2, and 2.3 in appendix 2, using display 2 to check that the results for the stress vector sP agree with your expectations.
Display 3: Stress Vector, Part 2
Stress vector components from matrix multiplication, stress vector envelope, normal and shear stresses acting on a plane
3.1 The stress vector components in terms of matrix multiplication
The appearance of stress calculations in three dimensions is simplified by rewriting the Cauchy formula in terms of a matrix multiplication. In two dimensions, the advantage of the matrix formulation is not so obvious but it is worthwhile to learn the method. We can take advantage of the equality of s xy and s yx to rewrite the Cauchy formula so that it looks like the multiplication of a two by two stress matrix with a two by one column matrix for the normal vector. In specifying the dimensions of a matrix, the first number gives the number of rows and the second number gives the number of columns in a matrix. The product of this matrix multiplication is the two by one column matrix for the stress vector.
We had for our fist Cauchy equation:
s Px=s xx*cos(a )+s yx*sin(a )
Now using PNx=cos(a ) , and PNy=sin(a ) for the x and y components of PN the vector normal to the plane and also using s yx=s xy , we get
s Px=s xx*PNx+s xy*Pny (1)
Similarly we had for our second Cauchy equation: s Py=s xy*cos(a )+s yy*sin(a ) from the Cauchy formula
Using the same substitutions for cos(a ) and sin(a ) we obtain
s Py=s yx*PNx+s yy*Pny (2)
When you recall the rules for matrix multiplication, you recognize the equations 1 and 2 are just the results of multiplying the two by two matrix defining the state of stress by the components of the normal vector to obtain the components of the stress vector acting on the plane perpendicular to the normal vector:
We can restate the above result in words: the stress vector for any plane is obtained by multiplying the stress tensor onto the vector normal to the plane.
The third display first shows the plane and the stress vector just as in the second display. By clicking on the box marked matrix in the control panel you can see the matrix multiplication displayed as well. You should confirm that you understand the elements in the matrix multiplication. Choose a simple matrix such as s xx=2 s yy=1 and s xy=0 for a start and do the multiplication for simple normal vectors like the normal vector along x and the normal vector along y. Then check out the multiplication for s xx=3, s yy=3, s xy=1 for the same normal vectors and finally for the normal making an angle of thirty degrees with the x axis.
You can also have the computer display the magnitude of the stress vector by clicking on the stress magnitude box in the control panel. Use display 3 to work exercise 3.1
3.2 The stress vector envelope
When you keep the stresses on the system fixed and vary the angle a , you find that the stress vector sP traces out an envelope. In display 3 click on the button "stress ellipse" in the control box. As you vary a , the corresponding stress vector is diplayed at the lower left always with its tail at the origin. Observe the curve or envelope that is traced out by the head of the stress vector. Exercises 3.2, 3.3, 3.4, and 3.5 in Appendix 2 provide suggestions for your explorations.
The envelope you have explored is sometimes called the stress ellipse. The longest, or major axis, is denoted by the symbol s1 and the shortest, or minor axis, by the symbol s3 and they are called the principal axes. Their lengths s 1 and s 3 are called the principal values and are shown in the figure.
In Ex 3.4 and 3.5, you should also have confirmed that the circle and the straight line are the limiting cases of the stress ellipse. In three dimensions the envelope becomes an ellipsoid, there are three principal axes all mutually perpendicular, and the corresponding principal values are denoted by s 1, s 2, and s 3 with s 1> s 2> s 3.
Display 4: Mohr circle, Part 1
Plot of s Ps vs s Pn in terms of s 1 and s 3
Display 4 lets you examine the normal and shear parts of the stress vector. The control panel is now simplified so that you choose the principal values s 1, s 3, and the angle of the normal a . A word of caution about varying the parameters s 1 and s 3. It is always assumed that s 1 is larger than or equal to s 3. The program prevents you from violating this rule. For instance if you have s 1=2 and you want to raise both s 1 and s 3 above 2, then you must raise s 1 first to its new value and then s 3, otherwise s 3 is limited above by the value of s 1. Similarly you must lower s 3 first when you want to decrease both parameters below the present value of s 3.
4.1 The normal and shear parts of the stress vector.
For many questions in the deformation or fracture of solids the values of the parts of the stress vector perpendicular and parallel to a plane are crucial in deciding whether distortion or motion will occur along that plane. The normal part sPn is just the part of the stress vector parallel to PN the normal to the plane. The normal component s Pn has magnitude equal to the length of the normal part and in geological applications the sign of the normal component is chosen to be positive for compression. The shear part sPs is the part of the stress vector parallel to the plane. The shear component s Ps has magnitude equal to the length of the shear part and the sign of the shear component s Ps is chosen to be positive when the shear part would produce a counterclockwise twist if it were applied a little outside the body. In our sketch below we have chosen an example in which both s Pn and s Ps are positive.
You should explore the behavior of the normal and shear parts of the stress vector with exercises 4.1, 4.2, 4.3, 4.4, and 4.5 in appendix 2. When you plot s Ps along y against s Pn along x you find that you trace out a circle which is usually called the Mohr circle.
In the exercises you should confirm that the center of the circle is on the horizontal axis at (s3+s1)/2 and that the radius is (s1-s3)/2. The angle g between the radius to the point at x=sPn and y=sPs and the horizontal as shown below, and in the exercises you should have confirmed that g=2a .
It is often important to relate the geometrical position of s Pn and s Ps on the Mohr circle to the actual orientation of the plane and the stress vector relative to the s1 direction. For instance, in discussing Display 5, we will find that we can determine the conditions for failure of a mechanical system by a geometrical comparison of the Mohr circle with a simple geometrical line showing the failure criterion. We can determine g from the position on the Mohr circle, and hence find a the direction of the normal to the plane. Thereforewe will be able to predict the plane along which failure will occur. As an example of the determination of a, consider a point on a Mohr circle with center at xc=2.0, s Pn=2.5, and s Ps=0.866 . (This point does not correspond to a failure condition). The angle g is found using tang=s Ps/(s Pn-xc)=.866/(2.5-2.0)=1.732 and g =60 degrees. Using a =g /2=30 degrees, we know that the normal PN makes an angle of 30 degrees with respect to the s 1 direction and we can draw the plane on which the stress vector acts. Finally we draw the parts s Pn with length 2.5 parallel to the normal direction and sPs with length .866 down to the right because of the positive sign. The stress vector sP is just the sum of sPn and s Ps as shown.
You can also find the values of s 1 and s 3 from the above data. The position of the center gives one relation for s 1 and s 3: xc=2.0=(s 1 + s 3)/2, and the distance of the point at s Pn and s Ps from the center gives the radius r of the Mohr circle and a second relation for s 1 and s 3. In our example, r=1=(s 1 - s 3)/2. So s 1=xc+r=3 and
s 3=xc-r=1. You can use
these values of s
1 and s 3 with a
=30 in display 4 to confirm the above figure and see the Mohr circle. Now
work exercise 4.6 in appendix 2.
Display 5 Mohr Circle, Part 2
Mohr circle and Failure Criteria, Cohesion, Envelope Slope
There is no simple law that lets you determine whether a system will fail if you are given the stresses on the system. However, Byerlee’s Rule (sometimes called Byerlee’s law) provides an approximate value of the stress that will produce failure: the system will not fail as long as the shear stress is less than the critical shear. The critical shear s cr is given by the prescription s cr=C+m*s N where the two parameters C and m are called the cohesion and the envelope slope and s N is just the magnitude of the normal stress. You can explore Byerlee’s rule by plotting the failure line (or for more complicated cases an envelope) along with the Mohr circle on a s Ps vs s Pn plot as shown in the figure.
The failure line is just a straight line on this plot. The cohesion C is the y intercept of the failure line and m is the slope of the failure line. The Mohr circle gives all the allowed values of s Ps and s Pn for the state of stress. As long as the Mohr circle lies below the failure line the system will not fail. When the stresses are changed so that the Mohr circle becomes tangent to the failure line, the system will fail and we can deduce a, the angle of the plane along which failure occurs, from the value g of the point where the Mohr circle just touches the failure line.
Display 5 shows the Mohr circle and the failure line. You can vary the cohesion C and the envelope slope m, as well as s 1, s 3, and the angle a that you used in display 4. In the upper right of display 5, we also show the curves giving the normal and shear stress components for all angles of the normal to the plane. The vertical purple line on the display marks the angle a selected. On the Mohr circle there also is a line from the center to the circle drawn at the angle g =2*a .
You should explore display 5 with exercises 5.1, 5.2, and 5.3 in appendix 2.
Once the program is started you can navigate in the various displays with use of the mouse. We will use "Click on" for moving the mouse pointer to a menu or check box selection and then depressing the left mouse key, and use "Click and Drag" for moving the mouse pointer to the small square in the scroll bars, depressing the left mouse key, holding it down and moving the small square until you have the desired value of the parameter.
You can choose a display by clicking on the display name. A new menu will appear at the top of the window, click on "Display" to go directly to work with the display, or click on "Back" to return to the main menu. When you have chosen the display, you can change the parameters of the display by using the scroll bars in the white dialog box. Click and drag the small squares in the scroll bars to make large changes in the parameters. For fine adjustments, click on the squares with the arrows at the ends of the scroll bars. For some of the displays, you can turn certain features on and off by clicking in the check boxes in the dialog boxes. To leave a display, click on stop in the dialog box and then on "Back" in the menu at the top of the window.
Finally, to stop the program, go back to the main menu and click on "Quit".
Appendix 2: Exercises
Display 1: Equilibrium
Ex 1.1 Using the control box, choose a simple set of forces by using s xx=2, s yy=0, and s xy=0. If you vary the width but not the height will L and R change? If you vary the height but not the width will L and R change. (in this as well as further exercises you will check you comprehension best if you make the prediction and calculate the result and then use the computer display to check you result) Is there any torque on the system?
Ex 1.2 Choose s xx=0, s yy=3, and s xy=0: predict the values of all the forces (and check your prediction with the display) for w=h=1; w=2,h=1; w=.5,h=2.
Ex 1.3 a Choose s xx=0, s yy=0, and s xy=2: and set w=2 and h=1. (note that on the stress diagram at the right s xy=s yx). Predict all the forces and calculate the torque about the center from each force and hence the total torque. b Now make h=.5 and compare the new torques with those in part a. This gives an experimental "proof" that with s xy=s yx the total torque always vanishes.
Ex 1.4 Now explore the range of possible values of s xx, s yy, s xy , w, and h.
a can you find a combination so that both L and B make an angle of 30 degrees with the x axis?
b can you find a combination so that L makes an angle of 30 degrees with the x axis and B makes an angle of 120 degrees with the x axis?
Display 2: Stress 1
Ex 2.1 Choose a triangular solid with a =30 degree and only s xx=2 different from zero as shown below. Before starting the program, make a sketch of the triangular solid with all the force vectors on the left and all the stress vectors on the right and sketch the stress vector sP. Check your result for sP against the Cauchy formulas. Then see whether your results agree with the program display.
Ex 2.2 Use the computer display and keep the same stress as in 2.1 but now let a go to 90 degrees. What happens to the area and the force on the right side? The behavior of FP and sP should agree with your expectations and you should again check the result with the Cauchy formula.
Ex 2.3 Now set s xx and s xy to zero but choose s yy=2. Predict the behavior for sP as a goes from zero to 90 and check your prediction with the display and by evaluating the Cauchy formula.
Display 3: Stress 2
Ex3.1 With s xx=s yy=3 and s xy=1, vary a and find the values of a for which the stress vector magnitude is a maximum and a minimum. Let b be the angle between the normal and the stress vector as shown in the sketch.
What are the values for b when the stress vector is a maximum and a minimum?
What is the angle in space between the stress vector at maximum and the stress vector at minimum? (These results are quite general and this process is often called "finding the principal axes" for the stress tensor).
Ex 3.2 Still with s xx=s yy=3 and s xy=1 vary a and draw the corresponding stress vector always with its tail at the origin. The head of the stress vector now traces out an envelope curve. Make a sketch of the envelope curve and check that your envelope includes the results for the maximum and minimum found in 3.1.
(You can confirm your sketch and also experiment with different values of s xx, s yy, and s xy by clicking on the "stress ellipse" in the control panel)
Ex 3.3 Now vary the values of the components of the stress tensor s xx, s yy, and s xy so that your envelope has the same size and shape as in 3.2 but now the longest (major) axis is along x and the shortest (minor) axis is along y.
Ex 3.4 Can you vary the stress tensor components s xx, s yy, and s xy so that your envelope becomes a circle of radius 2? A circle of radius 3? Is your result unique? (Try with s xy=0 and with s xy not equal to zero)
Ex 3.5 Can you vary the stress tensor components so the envelope becomes a straight line along the x axis?
Display 4: Mohr circle, part 1
Ex 4.1 Choose a simple state of stress such as s 1=3 and s 3=2. (This corresponds to choosing sxx=3, syy=1, and sxy=0 in part 3). Examine the display first when the stress vector sP is along the major (x) axis and then when sP is along the minor (y) axis of the stress ellipse.
What are the corresponding values of the normal and
Make a sketch of s Ps vs s Pn, that is plot the shear component s Ps in the up direction against the normal component s Pn in the horizontal direction for a =0, 45, 90, 135, and 180 degrees. (The values of s Ps and s Pn are displayed just to the right of plot of sP. ) Mark your points with the corresponding value of a . You may want to include some points in between to obtain a clear picture of the curve that is traced out. Your curve should be a circle as shown in the figure.
Ex 4.2 Leave s 1=3 and s 3=1 and click on the button marked Mohr circle to check your sketch in Ex 4.1. Now vary the angle a of the normal to the plane from 0 to 180 and examine the curve traced out on a s Ps vs s Pn plot. The resulting circle is called the Mohr circle. Find the rule for the location of the point furthest to the right and furthest to the left for your circle. Test your rule by choosing some new values for s 1 keeping s 3 fixed, and then choose new values for s 3 keeping s 1 fixed.
Ex 4.3 Return to some convenient values like s 1=3 and s 3=1 and now find the relationship between the angle a and the location of the s Pn, s Ps point on the circle. Call g the angle that the line from the center of the circle to the point at s Pn, s Ps makes with the horizontal s Pn axis. Determine g for a =0, 45, 90, 135, and 180 degrees and thereby "prove" that g =2*a .
Ex 4.4 Note that by setting a =0 you always choose the point furthest to the right (maximum normal stress) and for a =90 you always choose the minimum normal stress. With these two choices you can easily check the results of Ex4.1 and confirm the coordinates of the points on the s Ps vs s Pn plot are at s Pn =s 1, s Ps=0 and s Pn =s 3, s Ps=0. From this result find the formula for the location of the center and for the radius of the Mohr circle.
Ex 4.5 With the expressions found in Ex 4.4
a: find the values of s 1 and s 3 that yield a circle that goes through the origin and has radius 1.
b: find the values of s 1 and s 3 whose center is at 2 that has zero radius. What is the shape for the stress ellipse in this case? This is a two dimensional example of lithostatic stress. For lithostatic stress the magnitude of the stress is independent of the orientation of the plane, the stress is always perpendicular to the plane, and the shear component is always zero.
Ex 4.6 Given g
=90 degrees, s Pn=1.5,
and s Ps=1 draw the
corresponding point on a Mohr circle diagram i.e. with s
Pn along the x axis and s
Ps in the y axis direction. Now draw the Mohr circle using the fact that
=90 degrees. From your sketch calculate s
1, s 3, and a
and then confirm your results using display 4.
Display 5: Mohr circle, part 2
Ex 5.1 Set the cohesion to 0.3 and the envelope slope to 0.577 so that the envelope is at 30 degrees with the horizontal. With s 3=0.3 what value of s 1 will cause the system to fail? What is the angle a that the normal to the plane makes with s 1 just at failure? Is the value of s Ps the maximum allowed by the Mohr circle? (You can use the plot of s Ps vs a in the upper right to check your answer).
Ex 5.2 keep the cohesion and envelope slope fixed at 0.3 and 0.577 and increase s 3. Try to predict the changes of s 1 and a that will just produce failure. Now set s 3 to 0.6 and use the display to obtain the corresponding values for s 1 and a .
Ex 5.3 The simple straight "failure line" we have used in display 5 is a good approximation over a finite range of stresses. For a wide range of the principal stresses, the straight line is replaced by a more complicated failure envelope as shown.
For large values of the stresses the slope of this failure envelope is nearly zero. In this case, what angle a does the normal to the failure plane make with the s 1 direction? Is the value of s Ps now at the maximum allowed by the Mohr circle?
1 Stress on system sxx,sxy on left; syx, syy on bottom
2 Plane PL, normal to plane PN, stress vector sP acting on plane PL
3 Cauchy Equations
sPx= sxx*cos(a) + sxy* sin(a)
sPy= syx*cos(a) + syy*sin(a)
or with cos(a)=PNx and sin(a)=Pny
4 Stress ellipsoid: envelope of sP, principal axes s1, s3 ; (s1>s3)
5 Normal and shear components sPn , sPs
6 Mohr Circle plot of sPs vs sPn: xc=(s1+s3)/2; radius=(s1-s3)/2; tan g=sPs/(sPn-xc)
7 Failure envelope: y intercept=Cohesion, slope=envelope slope,
Mohr circle is tangent to envelope at failure.